Solve the differential equations (t - 1)3 ds/dt + 4(t

Gregory Jones

Gregory Jones

Answered question

2021-12-31

Solve the differential equations (t1)3dsdt+4(t1)2s=t+1,t>1

Answer & Explanation

censoratojk

censoratojk

Beginner2022-01-01Added 46 answers

Step 1
Given: (t1)3dsdt+4(t1)2 s=t+1;t>1
Step 2
Calculation:
(t1)3dsdt+4(t1)2s=t+1;t>1
dsdt+4(t1)2(t1)3s=t+1(t1)3
dsdt+4(t1) s=t+1(t1)3
Compare the differential equation with dydx+Py=Q
Here P=4(t1)2 andQ=t+1(t1)3
Solution of given differential equation is y.(I.F)=Q.(I.F)dx+c
Now, s.(t1)4=t+1(t1)3.(t1)4dt+c
s.(t1)4=(t+1).(t1)dt+c
s.(t1)4=(t21)dt+c
s.(t1)4=t33t+c
godsrvnt0706

godsrvnt0706

Beginner2022-01-02Added 31 answers

(t1)3dsdt+4(t1)2 s=t+1;t>1
(t1)3dsdt+4(t1)2s=t+1;t>1
dsdt+4(t1)2(t1)3s=t+1(t1)3
dsdt+4(t1) s=t+1(t1)3
dydx+Py=Q
P=4(t1)2 andQ=t+1(t1)3
y.(I.F)=Q.(I.F)dx+c
s.(t1)4=t+1(t1)3.(t1)4dt+c
s.(t1)4=(t+1).(t1)dt+c
s.(t1)4=(t21)dt+c
s.(t1)4=t33t+c
karton

karton

Expert2022-01-10Added 613 answers

Solution:
(t1)3dsdt+4(t1)2s=t+1;t>1
dsdt+4(t1)2(t1)3s=t+1(t1)3
dsdt+4(t1) s=t+1(t1)3
Compare the differential equation with dydx+Py=Q
P=4(t1)2 andQ=t+1(t1)3
Solution of given differential equation is y.(I.F)=Q.(I.F)dx+c
s.(t1)4=t+1(t1)3.(t1)4dt+c
s.(t1)4=t33t+c

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